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Math Help - Trignometry - modelling and problem solving

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    Trignometry - modelling and problem solving

    From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

    Please can you help with this problem
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Trignometry - modelling and problem solving

    From a point A due north of a tower, the angle of elevation to the top of the tower is 45 degrees. From point B, 100M on a bearing of 120 DEGREES FROM A the angle of elevation is 26 degrees. Find the height of the tower.

    Please can you help with this problem
    Trignometry - modelling and problem solving-castle_top.png

    From the image above Given: \angle EAD = 45^\circ, \angle BAD = 120^\circ and AB = 100m

    We have to find the height of the tower ED


    Draw from point G to CD, GH which is perpendicular to CD and to BE, GO perpendicular to BE


    In \triangle ABC:
    \begin{align*}\sin {\angle BAC} =& \frac{BC}{AB}\\ =& \frac{BC}{100}....\text{[because }AB = 100, \angle BAD = 120^\circ \text{ and }\angle BAC = 180^\circ - 120^\circ = 60^\circ\text{]}\end{align*}
    \therefore GH = BC = \frac{100\sqrt{3}}{2}


    In \triangle GAH:
    \sin {\angle GAH} = \frac{GH}{AG}
    \therefore AG = \frac{100\sqrt{3}}{\sqrt{2}}....\text{[} \angle GAH = 45^\circ \text{ and } GH = \frac{100\sqrt{3}}{2}}\text{]}


    In \triangle ABG:
    \begin{align*}BG =& \sqrt{AB^2 + AG^2 - 2*AB*AG*\cos {\angle BAG}}...\text{[ by cosine law]}\\=& 136.60...[AB = 100, AG = \frac{100\sqrt{3}}{\sqrt{2}}, \angle BAG = 75^\circ]\end{align*}


    In \triangle BOG:
    \sin{\angle OBG} = \frac{OG}{BG}
    \therefore OG = 59.88^\circ...[\angle OBG = 26^\circ \text{ and } BG = 136.60


    In \triangle BEF:
    \begin{align*}\angle FEB =& 180^\circ - \angle FBE - \angle BFE \\ =& 180^\circ - 26^\circ - 90^\circ\\=& 64^\circ\end{align*}
    Now \angle AGH = \angle GEF because GF is parallel to AD and straight line AE intersects them.


    In \triangle AGH:
    \angle AGH = 180^\circ - \angle GAH - \angle GHA = 45^\circ...[\angle GAH = 45^\circ, \angle GHA = 90^\circ


    Now:
    \angle FEB = \angle BEA + \angle AED

    Because \angle AED = \angle GEF = \angle AGH = 45^\circ\text{ and } \angle FEB = 64^\circ:

    \therefore \angle BEA = 19^\circ



    In \triangle OEG:
    \sin{\angle GEO} = \frac{OG}{GE}

    Because \angle BEA = 19^\circ \text{ and } OG = 59.88
    \therefore GE = 183.92



    In \triangle GEF:
    \cos{\angle GEF} = \frac{EF}{GE}

    But \angle GEF = 45^\circ \text{ and } GE = 183.92
    \therefore EF = 130.05


    Now:
    \begin{align*}ED = EF + FD = EF + BC =& 130.05 + \frac{100\sqrt{3}}{2}\\ =& 216.65\end{align*}

    So the height of the tower is 216.65 \text{ meter}
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